if set a ={2,3,5,7} and two numbers are selected at random

Question

if set a ={2,3,5,7} and two numbers are selected at random from A(duplicates allowed), then what is the probability that the sum of the numbers is even?

(A)1/8
(B)13/16
(C)5/12
(D)5/8
(E)2/3

chetan2u · Answer

Hi,

there are 3 odd numbers and 1 even number..
the way Sum can be found is 4*4 = 16 as same number can be picked up again..

1) even number cannot pair up with any other number to have EVEN sum, it can just pair up with itself so (2,2) = 1 way..
2) the 3 odd can be paired with themselves or any other odd number to produce EVEN sum = 3*3 =9 
total =  1+9 = 10

Prob = \frac{10}{16}= \frac{5}{8} 
D

adiagr · Answer

sum of the two numbers would be even if 
(1) Both digits are even.
(2) Both digit are odd.

(1) only one case is there: 22
(2) Following cases are there: 33, 35, 37 - 3 nos. Similarly 3 nos, with 5 (53, 55, 57) and 3 nos. with 7 (73, 75, 77)

Total 10 nos.

Total nos. that can be formed (repetition is allowed)

{1st no. can be chosen in 4 ways. Similarly 2nd no. can be chosen in 4 ways. Total 4*4 = 16 ways}

Prob : 10/16 =  5/8

D is the answer.

Divyadisha · Answer

Total possible combinations= 4*4

Possibility of having odd numbers= if one number is even and one is odd

Let's say 1st number is 2 and second can be chosen from remaining 3 and vise versa= 1*3*2

Probability of choosing odd numbers= 3*2/4*4= 3/8

Probability of choosing even number= 1-3/8= 5/8

D is the answer

msk0657 · Answer

For this question, we can do very easily without calculation.

Given that two numbers are selected and the numbers can be duplicates - so total possible cases are 4 * 4 ( any four numbers * any four numbers).

Given that sum of the numbers is even.
So possible cases are ( 2,2 ) (3,3) (5,5) (7,7) with same numbers. => 4 cases.

Adding 2 with other number will not give even sum.
Now (3,5) (3,7) 
 (5,3) (5,7)
 (7,3) (7,5) => 6 cases

Now total will be 10 cases.

P = number of possible cases / Total cases.
 => 10 / 16 => 5/8 --> Option D.

snorkeler · Answer

msk0657  Infact finding an odd combination is easy rather than even, and then subtracting from 1. (Only risk involved is one might forget to subtract) 
 Divyadisha  solved using this approach

This question is pretty straight forward so one might calculate either P(even) or 1-P(odd) is fine.
Some questions, in which multiple scenarios are needed to calculate the probability then (1-n) method will be handy in most of the cases. 
"Atleast" questions are best solved using (1-n) method.

rocky710 · Answer

Hi to all,

First of all thank you for your valuable contribution to the discussion.

Nevertheless, I do not agree with you.

The question is asking for two numbers selected at random, NOT a two-digit number composed of two numbers selected at random from set A. Therefore, it would be indifferent to select (5,7) or (7,5) as the sum of these two pairs is the same. In this case, I would solve the problem using Combinatorics instead of Permutations and I would get  7/10  as a result.

Please correct me if I'm wrong as I'm still struggling with this particular question.

Thank you.

chetan2u · Answer

If you are picking two at one time or randomly, you will not get duplicates. 
Duplicate means that one number is picked, then placed back and again picked. So order is important.

rocky710 · Answer

Oh I see it now. Thank you so much!!

ScottTargetTestPrep · Answer

Solution:

We can obtain an even sum of two numbers by getting either two even numbers or two odd numbers.

P(an even sum)

= P(even, even) + P(ood, odd)

= 1/4 x 1/4 + 3/4 x 3/4

= 1/16 + 9/16

= 10/16 = 5/8

Answer: D

Kinshook · Answer

Asked: If set A ={2,3,5,7} and two numbers are selected at random from A(duplicates allowed), then what is the probability that the sum of the numbers is even?

Since only 2 is even and 3,5,7 are odd.
Sum is even when either 2 numbers are even or 2 numbers are odd. 
Probability = P(even)*P(even) + P(odd)*P(odd) = 1/4 * 1/4 + 3/4 * 3/4 = 1/16 + 9/16 = 10/16 = 5/8

IMO D

Basshead · Answer

To get an even sum the two integers need to be (even, even) or (odd, odd)

P(even, even) = 1/4 * 1/4 = 1/16
P(odd, odd) 3/4 * 3/4 = 9/16

1/16 + 9/16 = 10/16 = 5/8

bumpbot · Answer

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if set a ={2,3,5,7} and two numbers are selected at random

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